It's a system of equations word problem

Question

The number 1 is subtracted from the first number, the difference is squared, then doubled; the result is equal to the second number. The number 1 is added to the first number, and the sum is squared; the result is equal to 4 minus the second number. a) create a system of equations to represent this relationship. B) solve the system to determine the numbers.

Answer 1

x=first number

y=second number

 

2(x-1)^2=y

(x+1)^2=4-y

substitute

(x+1)^2=4-2(x-1)^2

x^2+2x+1=4-(2[x^2-2x+1])

x^2+2x+1=4-2x^2+4x-2

x^2+2x+1=2-2x^2+4x

x^2+2x+1-2+2x^2-4x=0

3x^2-2x-1=0

factor

(3x+1)(x-1)=0

3x+1=0

3x=-1

x=-1/3

substitute into the first equation to get y

2(-1/3-1)^2=y

2(-4/3)^2=y

2(16/9)=y

y=32/9

(-1/3, 32/9) is one solution

 

x-1=0

x=1

substitute into the first equation to get y

2(1-1)^2=y

2(0)=y

y=0

(1, 0) is the second solution

Answer 2

First step in this problem is to assign variables for the numbers. Let us call the first number x and the second number y.

 

According to the problem stated, the number one is subtracted from x, and then that difference is squared. This can be written as (x-1)². Then you double it, which is the same as saying you're multiplying that by 2. This will be equal to the number y. This is your first equation in a system of 2 equations.

 

The second equation is set up similarly. 1 is added to x, then that quantity is squared, i.e. (x+1)². This is equal to 4 minus y.

 

At this point in time, now that you have two equations and two unknowns, you can solve either using substitution, since the first equation is equal to y, or you can use cancellation, which is essentially the same thing as substitution. This will allow you to solve for one variable at a time, at which point you can plug this value into either equation to determine the remaining variable. This should get you to the end of the problem.

 

If you need any more help or clarification, please feel free to reach out to me.